Question
In which of the following pairs are both the ions coloured in aqueous solution?
In which of the following pairs are both the ions coloured in aqueous solution?
$$\left( {{\text{At}}{\text{. no}}{\text{.}}\,Sc = 21,Ti = 22,} \right.$$ $$\left. {Ni = 28,Cu = 29,Co = 27} \right)$$
A.
$$N{i^{2 + }},T{i^{3 + }}$$
B.
$$S{c^{3 + }},T{i^{3 + }}$$
C.
$$S{c^{3 + }},C{o^{2 + }}$$
D.
$$N{i^{2 + }},C{u^ + }$$
Answer :
$$N{i^{2 + }},T{i^{3 + }}$$
Solution :
$${}_{28}Ni = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^8},4{s^2}$$
$$N{i^{2 + }} = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^8}$$
$$3{d^8}$$
( 2 unpaired electrons)
$${}_{22}Ti = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^2},4{s^2}$$
$$T{i^{3 + }} = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^1}$$ (1 unpaired electron)
$${}_{21}Sc = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^1},4{s^2}$$
$$S{c^{3 + }} = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}$$ (no unpaired electron)
$${}_{29}Cu = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}},4{s^1}$$
$$C{u^ + } = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}}$$ (no unpaired electron)
Hence, in the above ions, $$N{i^{2 + }}$$ and $$T{i^{3 + }}$$ are coloured in aqueous solution due to the presence of unpaired electrons in d subshell.
$${}_{28}Ni = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^8},4{s^2}$$
$$N{i^{2 + }} = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^8}$$
$$3{d^8}$$
( 2 unpaired electrons) $${}_{22}Ti = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^2},4{s^2}$$
$$T{i^{3 + }} = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^1}$$ (1 unpaired electron)
$${}_{21}Sc = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^1},4{s^2}$$
$$S{c^{3 + }} = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}$$ (no unpaired electron)
$${}_{29}Cu = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}},4{s^1}$$
$$C{u^ + } = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}}$$ (no unpaired electron)
Hence, in the above ions, $$N{i^{2 + }}$$ and $$T{i^{3 + }}$$ are coloured in aqueous solution due to the presence of unpaired electrons in d subshell.